regex - How can I extract a percentage number without the "%" symbol from a string?

Question

-InternalBattery-0 (id=7405667) 100%; charged; 0:00 remaining present: true

I want to do egrep -ow (regex) on the above string and get only 100 of 100% and nothing else. How can I do that?

标签： regexbashshellregex-lookaroundsregex-group

Answer 1

sed with a backreferece to reinsert the digits before the percent sign in the general substitution form is probably a bit more doable than grep. For example you can use:

sed -E 's/^[^)]+\)\s+([0-9]+)%.*$/\1/'

Which matches one of more characters from the beginning that are not ')', then a literal ')' and then matches any amount of whitespace. A capture group begins capturing digits up to the next '%' with remaining characters to end of line discarded. The first backreference \1 is used to replace the whole line with what was captured between the (...) in ([0-9+)%.

Example Use/Output

$ echo "-InternalBattery-0 (id=7405667) 100%; charged; 0:00 remaining present: true" | 
sed -E 's/^[^)]+\)\s+([0-9]+)%.*$/\1/'
100

Awk Solution

Since in the comment you mention you are attempting this in Apple Script (which I know little about), then perhaps a straight-forward awk solution that simply loops to find the field that contains the '%' character and than chops-off everything from '%' to the end of the field, prints the result and exits, e.g.

$ echo "-InternalBattery-0 (id=7405667) 100%; charged; 0:00 remaining present: true" | 
awk '{ for( i=1; i<=NF; i++) if ($i ~ /%/) { sub(/%.*$/,"",$i); print $i; exit }}'
100

That way you can, again, use the '%' character to identify the proper field in your string of text and then simply remove it, and anything that follows, and be confident that you have the correct result if '%' only appears once in your line of input.

Give it a try and let me know if you have further questions.